Code & Fun

第67天。

今天的题目是Valid Triangle Number:

Given an array consists of non-negative integers, your task is to count the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.
Example 1:
Input: [2,2,3,4]
Output: 3
Explanation:
Valid combinations are:
2,3,4 (using the first 2)
2,3,4 (using the second 2)
2,2,3
Note:
The length of the given array won’t exceed 1000.
The integers in the given array are in the range of [0, 1000].

莫名其妙的用一个O(n^3)的解法AC了:

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int triangleNumber(vector<int>& nums) {
sort(nums.begin(),nums.end());
int ret = 0;
for(int i = 0;i < nums.size();i++) {
for(int j = i + 1;j < nums.size();j++) {
//nums[i] + num[j] > a;
for(int k = j+1;k < nums.size() && nums[i] + nums[j] > nums[k];k++)
ret++;
}
}
return ret;
}

然后是dicuss中的O(n^2)的解法:

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public static int triangleNumber(int[] A) {
Arrays.sort(A);
int count = 0, n = A.length;
for (int i=n-1;i>=2;i--) {
int l = 0, r = i-1;
while (l < r) {
if (A[l] + A[r] > A[i]) {
count += r-l;
r--;
}
else l++;
}
}
return count;
}

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